# Difference between revisions of "1953 AHSME Problems/Problem 8"

## Problem 8

The value of $x$ at the intersection of $y=\frac{8}{x^2+4}$ and $x+y=2$ is:

$\textbf{(A)}\ -2+\sqrt{5} \qquad \textbf{(B)}\ -2-\sqrt{5} \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \text{none of these}$

## Solution

$x+y=2\implies y=2-x$. Then $2-x=\frac{8}{x^2+4}\implies (2-x)(x^2+4)=8$. We now notice that $x=0\implies (2)(4)=8$, so $\fbox{C}$